What is the #101#st derivative term in the Taylor series of #tan^(-1)(x)#?

Probably the best way to find this is to construct the Taylor series centered at #x=0# for #tan^{-1}(x)# and use the pattern to find the appropriate coefficient, and then, ultimately, the value of #f^{(101)}(0)# (the value of the #101#st derivative of #f# at #x=0#).

The best way to find the Taylor series for #tan^{-1}(x)# centered at #x=0# is to integrate the series for #1/(1+x^2)# (centered at #x=0#) term-by-term.

The expression

#1/(1+x^2)=1/(1-(-x^2))#

is the sum of a geometric series with firsts term #a=1# and common ratio #r=-x^2#. Hence,

#1/(1+x^2)=1-x^2+x^4-x^6+x^8-x^10+cdots#

(for #|x|<1#). Since #tan^{-1}(0)=0#, it follows that

#tan^{-1}(x)=int 1/(1+x^2)dx#

# =x-x^3/3+x^5/5-x^7/7+x^9/9-x^11/11+cdots#

(for #|x|<1#...and actually, it works when #x=1# as well!)

By the pattern and the general formula for Taylor series, it follows that #f^{(101)}(0)/(101!)=1/101#.

Therefore,

#f^{(101)}(0)=(101!)/101=100!#.

If #tan^(-1)(x)# is interpreted to mean the #arctan(x)#
and assuming #arctan# is restricted to the range #(-pi/2,pi/2]#

Then #f(0) = arctan(0) = 0#
and
#f^101(0) = 0^101 = 0#